0
$\begingroup$

$$\lim_{x \to +0} \sin{x} \cdot \ln{(\sin{x})}$$

How would you do this without L'Hospital's rule or using a table of values? A step by step explanation would be nice.

  • 0
    I'm not sure where to begin. – 2017-02-27
  • 1
    Use substitution: $u=\sin x$,and you'll get $\lim_{u\to0^+} u\ln u$, which is a high-school limit. – 2017-02-27
  • 1
    Substitute $t=\sin x$. Then $t\to 0^+$ as $x\to 0^+$. So, the limit $$\lim\limits_{t\to 0^+}t\ln t$$ is needed to compute. Using L'Hospital rule we arrive at $0$. Then we know the limit, it is enough to show that is is really zero. – 2017-02-27
  • 1
    The last one is a standard limit. L'Hospital is not required. – 2017-02-27
  • 0
    Thank you! Is there a way to prove this standard limit, without using l'hopitals rule?? – 2017-02-27

2 Answers 2

2

First of all, let's deal with the elephant in the room: for $x>0$ small enough, since $\frac{\sin x}{x} \xrightarrow[x\to0]{} 1$ we have $\sin x >0$, and therefore $\ln \sin x$ is well defined on an interval of the form $(0,\varepsilon)$.

Now, the limit.

  • A simple solution would be to use Taylor approximations: since $\sin x = x+o(x)$ when $x\to 0^+$, we have $$ \sin x \cdot \ln\sin x = (x+o(x))\ln(x+o(x)) = (1+o(1))(x\ln x+\ln(1+o(1)) \xrightarrow[x\to0^+]{} 0 $$ since $\lim_{x\to 0^+}x\ln x = 0$ and $\ln(1+o(1)) \xrightarrow[x\to0^+]{} \ln 1= 0$.

  • Another solution, even simpler, is to argue that $u\stackrel{\rm def}{=}\sin x\xrightarrow[x\to0^+]{}0$, and so $$ \sin x \cdot \ln\sin x = u\ln u \xrightarrow[x\to0^+]{}0 $$ since $\lim_{x\to 0^+}x\ln x = 0$.

  • 0
    thank you. Also, is there a way to prove xlnx goes approaches 0, without using l'hopitals rule. – 2017-02-27
  • 0
    @A.AK Yes -- see e.g. [this](http://math.stackexchange.com/questions/258273/how-do-i-prove-that-lim-x%E2%86%920-x%E2%8B%85-ln-x-0). – 2017-02-27
  • 0
    To the downvoter: care to explain why? – 2017-02-28
0

Alternative way to show that $\lim \limits_{x \rightarrow 0^+} x \ln x = 0$: Differentiate $f(x) = x \ln x$ to see that it is decreasing in a right neighbourhood of $0$. To show that all decreasing sequences $(x_n)$ with limit $0$ will give $\lim f(x_n) = 0$ it is then enough to find the limit for just one such sequence, by monotonicity of $f$. It's easy to see that $e^{-n}\ln(e^{-n})=-ne^{-n} \rightarrow 0$.